Simplify and expand the following expression: $ \dfrac{3}{5y - 40}- \dfrac{4}{2y - 14}- \dfrac{2}{y^2 - 15y + 56} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{3}{5y - 40} = \dfrac{3}{5(y - 8)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{4}{2y - 14} = \dfrac{4}{2(y - 7)}$ We can factor the quadratic in the third term: $ \dfrac{2}{y^2 - 15y + 56} = \dfrac{2}{(y - 8)(y - 7)}$ Now we have: $ \dfrac{3}{5(y - 8)}- \dfrac{4}{2(y - 7)}- \dfrac{2}{(y - 8)(y - 7)} $ The least common multiple of the denominators is: $ 10(y - 8)(y - 7)$ In order to get the first term over $10(y - 8)(y - 7)$ , multiply by $\dfrac{2(y - 7)}{2(y - 7)}$ $ \dfrac{3}{5(y - 8)} \times \dfrac{2(y - 7)}{2(y - 7)} = \dfrac{6(y - 7)}{10(y - 8)(y - 7)} $ In order to get the second term over $10(y - 8)(y - 7)$ , multiply by $\dfrac{5(y - 8)}{5(y - 8)}$ $ \dfrac{4}{2(y - 7)} \times \dfrac{5(y - 8)}{5(y - 8)} = \dfrac{20(y - 8)}{10(y - 8)(y - 7)} $ In order to get the third term over $10(y - 8)(y - 7)$ , multiply by $\dfrac{10}{10}$ $ \dfrac{2}{(y - 8)(y - 7)} \times \dfrac{10}{10} = \dfrac{20}{10(y - 8)(y - 7)} $ Now we have: $ \dfrac{6(y - 7)}{10(y - 8)(y - 7)} - \dfrac{20(y - 8)}{10(y - 8)(y - 7)} - \dfrac{20}{10(y - 8)(y - 7)} $ $ = \dfrac{ 6(y - 7) - 20(y - 8) - 20} {10(y - 8)(y - 7)} $ Expand: $ = \dfrac{6y - 42 - 20y + 160 - 20}{10y^2 - 150y + 560} $ $ = \dfrac{-14y + 98}{10y^2 - 150y + 560}$ Simplify: $ = \dfrac{-7y + 49}{5y^2 - 75y + 280}$